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  • HDU 1041 Computer Transformation(高精度)

    Computer Transformation

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
    Total Submission(s): 3845    Accepted Submission(s): 1420


    Problem Description
    A sequence consisting of one digit, the number 1 is initially written into a computer. At each successive time step, the computer simultaneously tranforms each digit 0 into the sequence 1 0 and each digit 1 into the sequence 0 1. So, after the first time step, the sequence 0 1 is obtained; after the second, the sequence 1 0 0 1, after the third, the sequence 0 1 1 0 1 0 0 1 and so on.

    How many pairs of consequitive zeroes will appear in the sequence after n steps?
     
    Input
    Every input line contains one natural number n (0 < n ≤1000).
     
    Output
    For each input n print the number of consecutive zeroes pairs that will appear in the sequence after n steps.
     
    Sample Input
    2 3
     
    Sample Output
    1 1
     
    Source
     
    Recommend
    JGShining
     
     
    先找到规律,推公式。
    1->01 ,  0->10
     
    而且很容易知道连续的0肯定是两个连续的0.
     
    设f[n]为n步操作后连续0的个数
     
    则连续的0怎么样来呢?只能由上一层的01变成,也就是上一层的01一定可以产生连续00
    上一层的01可以由再上一层的1得到。或者由上一层的00也可以产生一个01.
    所以递推公式产生了:
    f[n]=f[n-2]+2^(n-3).
     
    由这个递推公式很容易产生通项公式:
    当n为偶数时,f[n]=(2^(n-1)+1)/3;
    当n为奇数时,f[n]=(2^(n-1)-1)/3;
     
    所有用大数公式就得出来了。。
     
    用JAVA写大数写出来的。。
    /*
     f[n]=f[n-2]+2^(n-3);
     
     
     
     n为奇数时,f[n]=(2^(n-1)-1)/3;
     n为偶数时,f[n]=(2^(n-1)+1)/3;
     
     
     
     */
    
    
    import java.util.*;
    import java.math.*;
    import java.io.*;
    public class Main {
        public static void main(String[] args) {
            Scanner cin=new Scanner(new BufferedInputStream(System.in));
            BigInteger a[]=new BigInteger[1000];
            a[0]=BigInteger.valueOf(1);
            for(int i=1;i<1000;i++)
                a[i]=a[i-1].multiply(BigInteger.valueOf(2));
            int n;
            BigInteger ans;
            while(cin.hasNextInt())
            {
                 n=cin.nextInt();
                 if(n%2==0)//偶数
                 {
                     ans=a[n-1].add(BigInteger.valueOf(1));
                     ans=ans.divide(BigInteger.valueOf(3));
                 }
                 else
                 {
                     ans=a[n-1].subtract(BigInteger.valueOf(1));
                     ans=ans.divide(BigInteger.valueOf(3));
                 }
                 System.out.println(ans);
            }
    
        }
    
    }
    import java.util.*;
    import java.math.*;
    import java.io.*;
    public class Main {
        public static void main(String[] args) {
            int n;
            Scanner cin=new Scanner(new BufferedInputStream(System.in));
            BigInteger a=BigInteger.valueOf(2);
            BigInteger ans;
            while(cin.hasNextInt())
            {
                n=cin.nextInt();
                if(n%2==1)
                  ans=a.pow(n-1).subtract(BigInteger.valueOf(1)).divide(BigInteger.valueOf(3));
                else
                  ans=a.pow(n-1).add(BigInteger.valueOf(1)).divide(BigInteger.valueOf(3));
                System.out.println(ans);
            }
    
        }
    
    }

    辛辛苦苦C++写了用份string的高精度。。竟然超时了。。。。

    效率不高

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  • 原文地址:https://www.cnblogs.com/kuangbin/p/2634090.html
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